The purpose of this question is to remind students that the standard trigo ratios (i.e. TOA CAH SOH -> Tan: Opp/Adj; Cos: Adj/Hypo; Sin: Opp/Hypo) applies only to an *acute* angle.

In order to solve for angle DCB which is an *obtuse* angle in the question, students need to remember a further set of equations i.e:

sin θ = sin (180^{o} - θ)

cos θ = -cos (180^{o} - θ)

tan θ = -tan (180^{o} - θ)

Hence exam questions usually require multiple steps like these. If everyday just simply TOA CAH SOH TOA CAH SOH, the examination board can close shop *liao*!

So remember this well!

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